I understand R1 & R2 are for feedback which the regulator uses to set output voltage. What is the purpose of the 50k ohm resistor, BJT, and the lower 10uF cap? Does the capacitor & resistor make some time constant the affects the feedback voltage?

If I may ask a semi-related question about the diode... I presume it's some sort of protection diode but what is it actually protecting?

NCP1117 slow turn-on

  • \$\begingroup\$ If your presumption is right about the diode, it could have been used to clamp negative voltages from inductive kickback. \$\endgroup\$
    – tangrs
    Jun 26, 2015 at 3:25
  • 2
    \$\begingroup\$ @tangrs no, the datasheet explains it fully. There is an internal feedback protection diode but it's only good for about 150mA. If the output is shorted and there's a sufficiently high capacitance in the feedback circuit the regulator could be damaged as it discharges through the internal diode. The external diode is there to take the brunt of that current draw. \$\endgroup\$
    – akohlsmith
    Jun 26, 2015 at 3:49
  • \$\begingroup\$ "A diode is placed between C1 and VOUT to provide a path for the capacitor to discharge." www.ti.com/lit/ds/symlink/lm117.pdf#page=17 \$\endgroup\$
    – endolith
    Aug 30, 2016 at 18:14

3 Answers 3


Image from datasheet

The reference voltage is programmed to a constant current source by resistor R1, and this current flows through R2 to ground to set the output voltage

\$I_{adj}\$ gradually falls down as the PNP turns off(in OP Figure) after charging the capacitor. The regulator initially observes that \$V_{out}\$ is actually higher (as \$I_{adj}\$ is higher) as per the equation above and hence will try to reduce the \$V_{out}\$. This in turn creates the slow turn on function as \$V_{out}\$ slowly rises as \$I_{adj}\$ gradually falls to minimum.


Initially, the capacitor voltage is zero and the PNP transistor is on, shorting out R2 and fixing the output at ~1.25V (see the datasheet on how R1/R2 set the output voltage). Since there is still some output voltage, it will slowly charge the base-collector capacitor through the 50k resistor, bringing the transistor base-emitter voltage down until it finally shuts off completely. Now R2 is pretty much the only element in the feedback loop, giving the desired output.

The diode is there to protect the regulator from cases where the output is shorted. It's described in the "protection diodes" section of the datasheet.

It's an interesting little circuit.


Here are the calculations I did using Excel and assuming: Vin=18V Vout=15v R1=120-Ohms R2=1320-Ohms (So that Vou=1.25*(1+R2/r1) gives us Vout=15V

A> at first the PNP Tranny shorts the Current at ADJ to ground and Vout is 1.25V (Because the Vout is created larger due to the constant current made by the Resistor between Vout and ADJ and the Voltage Drop across it always being 1.25V).

B> There is a current at ADJ and I assumed that it would flow through the Parallel resistors +the Emitter of the PNP accordingly

C> The Parallel Resistance creates a normal Voltage drop of 12.86 across R2||50k

D> Assumed that the PNP had a beta of 100 and the Current through the 50k-ohm resistor was 12.86/50000=.000264A making the remaining current at the ADJ node shunted to GND through the PNP.

E> The Capacitor will Charge at an RC time constant of .00001*50000=.5 seconds and so it will reach a stored voltage (plus the Vbe of 0.7 between the Bass and the emitter) exceeding 12.83V by the time 1 second passes and the PNP Turns off just at 1 seconds

F> After one second the Capacitor stores it's Voltage and keeps the PNP out of the circuit until the Circuit is turned off and the Diode drains the Cap through the Load or through the 50k-ohm resistor and R2

Maybe this is what is going on?

enter image description here


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